Undervoltage Battery Dectection Circuit

[Cazality]

Hi everyone,

I just came across this site while doing some research on the circuit I am trying to build and found it to be full of information with lots of interesting projects.

I am trying to build an under voltage detection circuit for a 12V car battery. The vehicle this would go into uses several pieces of equipment and recently people have been forgetting to turn off the equipment when they are finished, and thus the battery is dead the next day.

I want to build the circuit using an op amp as my voltage comparator and have attached a picture of what I have so far. I would like the circuit to turn off everything when the voltage reaches about 12.2V. I have attempted to do that using a 6.2 zenor diode as my reference voltage and a voltage divider for my variable voltage.

I also wanted to add in a reset switch that someone would need to push in order to turn the system back on, but I am not sure how to implement that yet.

There is still several issue with this setup so far I think with the transistor and all, but I was hoping I could get some feedback and suggestions on what I have so far.

Sorry for the long post.

Thanks for your help!

 

[crutschow]

You should have a resistor at the op amp output to the base of the transistor, and some supply voltage for the collector resistor.

You show two batteries for op amp power. I assume you just have one battery (the vehicle battery) to power everything. The op amp negative power terminal should thus go to ground. The positive terminal should have a small series resistor and a capacitor to ground to filter the voltage and suppress the high voltage spikes that can occur during vehicle starting.

If I understand correctly, for the reset you could just connect it to the ignition switch such that when the engine is running (and generator charging), the monitor circuit is not active. For example connecting the ignition voltage to the junciton of R1 and R2 though a diode will make the circuit think the battery voltage is high when the ignition is on.

12.2V might be a little low to shut off the circuits. This leaves less than 40% of the charge which may be marginal to start the engine. Recomment more like 12.4 to 12.5V which would leave about 60 to 70%.

 

[Boncuk]

12.2V might be a little low to shut off the circuits. This leaves less than 40% of the charge which may be marginal to start the engine. Recomment more like 12.4 to 12.5V which would leave about 60 to 70%.

Even 10.5V is a good value for shut-off. I used that trip point to cut off power of the windshield heating with no problems for a cold engine start. (Ford 2.3l 6-cyl V-engine) The battery voltage drops to 12.5 if the battery isn't being charged. In that case he might save the money for an undervoltage detector.

 

[Cazality]

Thanks for the responses and suggestions everyone. I am working on making the changes you suggested, and I fixed my circuit so everything is being powered by a single supply and added the filter to the positive supply of the op amp.

I have one question about the collector being powered on the transistor. Should I attach it through a diode to the positive power rail and then have my output going to a relay of some sort? I am slightly confused by how the circuit would turn off the equipment if the collector is always powered.

I am probably just missing something. I had figured when the op amp was "high" current would flow through the collector and to the output and therefore the equipment would have power. When the op amp goes "low" current through the collector is cut off, and therefore cuts off the output (equipment).

Thanks again, appreciate it.

 

[ericgibbs]

I have one question about the collector being powered on the transistor. Should I attach it through a diode to the positive power rail and then have my output going to a relay of some sort? I am slightly confused by how the circuit would turn off the equipment if the collector is always powered.

I am probably just missing something. I had figured when the op amp was "high" current would flow through the collector and to the output and therefore the equipment would have power. When the op amp goes "low" current through the collector is cut off, and therefore cuts off the output (equipment).

Thanks again, appreciate it.

hi,
The transistor will have to drive a relay, the relay contacts will switch the power to the external equipment.

 

[crutschow]

I have one question about the collector being powered on the transistor. Should I attach it through a diode to the positive power rail and then have my output going to a relay of some sort? I am slightly confused by how the circuit would turn off the equipment if the collector is always powered.

Yes, if you have significant current to control you would probably want a relay. One wire of the relay coil would be tied to the +12V and the other to the collector so the relay would be on when the transistor is conducting. The transistor must be large enough to carry the relay coil current.

The transistor base resistor's value should be selected to give about 1/10 of the collector (coil) current into the transistor base. The typical base-emitter voltage when the transistor is on is about 0.65V.

Be sure to add a diode across the relay coil to absorb the inductive kick and avoid zapping the transistor. The diode cathode should be connected to the +12V side of the coil and the anode to the collector side of the coil.

 

[Cazality]

Thanks again guys. When I get a completed design I will post it here.

 

[crutschow]

Even 10.5V is a good value for shut-off. I used that trip point to cut off power of the windshield heating with no problems for a cold engine start. (Ford 2.3l 6-cyl V-engine) The battery voltage drops to 12.5 if the battery isn't being charged. In that case he might save the money for an undervoltage detector.

You might want to combine R1 and R2 into a trimpot with wiper going to the op amp input. That way you can adjust the trip point the exact value you need and also compensate for component tolerances.